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In game theory, " guess 2 3 of the average " is a game that explores how a player’s strategic reasoning process takes into account the mental process of others in the game. [1] In this game, players simultaneously select a real number between 0 and 100, inclusive. The winner of the game is the player (s) who select a number closest to ...
Also bear in mind that the fraction 2/3 is the single exception, used in addition to integers, that Ahmes uses alongside all (positive) rational unit fractions to express Egyptian fractions. The 2/n table can be said to partially follow an algorithm (see problem 61B) for expressing 2/n as an Egyptian fraction of 2 terms, when n is composite.
17 indivisible camels. The 17-animal inheritance puzzle is a mathematical puzzle involving unequal but fair allocation of indivisible goods, usually stated in terms of inheritance of a number of large animals (17 camels, 17 horses, 17 elephants, etc.) which must be divided in some stated proportion among a number of beneficiaries.
In terms of partition, 20 / 5 means the size of each of 5 parts into which a set of size 20 is divided. For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is ...
In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator. [ 1]
In mathematics, a harmonic progression (or harmonic sequence) is a progression formed by taking the reciprocals of an arithmetic progression . Equivalently, a sequence is a harmonic progression when each term is the harmonic mean of the neighboring terms. As a third equivalent characterization, it is an infinite sequence of the form.
The misdirection in this riddle is in the second half of the description, where unrelated amounts are added together and the person to whom the riddle is posed assumes those amounts should add up to 30, and is then surprised when they do not — there is, in fact, no reason why the (10 − 1) × 3 + 2 = 29 sum should add up to 30.
If two divisions are done, a multiple of 5 · 5=25 rather than 5 must be used, because 25 can be divided by 5 twice. So the number of coconuts that could be in the pile is k · 25 – 4. k =1 yielding 21 is the smallest positive number that can be successively divided by 5 twice with remainder 1.
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